Electronic – How to calculate the node voltages for this circuit using the voltage divider rule

circuit analysisvoltage divider

I am working on the following circuit:
enter image description here

Simplifying the circuit:
enter image description here

After simplifying, my answers do not match that of values I get from simulating the circuit in multisim. Here's my calculation:
$$Voltage V_B$$
$$V_B=6(\frac{5000}{5000+(32.35714286\times10^3)})$$
$$V_B=0.8030592734V$$
$$6-0.8030592734=5.20V$$

$$Voltage Vc$$
$$V_c=5.2(\frac{(32.35714286\times10^3)}{5000+(32.35714286\times10^3)})$$
$$V_c=4.50V$$

The simulations I did on multisim show that the voltages for Vb and Vc are:
$$V_B=5.161V$$
$$V_c=5.035V$$

I don't know if its the simplification I'm doing wrong or my approach towards the voltage divider rule. I need help.

Best Answer

Your first circuit has a resistor between C and GND. You seem to have added it to your bottom resistor. You also seem to have your 1k||3k resistor calculation incorrect. 3k*1k/3k+1k is 750 ohms. Thus you should end up with a circuit like this:

schematic

simulate this circuit – Schematic created using CircuitLab

From here, it is simpler to do Ohms Law to find the current in the circuit, then find the voltage drops of all the resistors.

If you must use the voltage divider rule, then you need to know if you are finding the voltage drop(s) across the resistor(s), or the voltage at the points with respect to GND, because that will make a difference to how you calculate it.

Of course, you always have the option to simplify to 2 resistors with the R2 component in the voltage divider as (R2+R3) too. It depends what your task is.

Another thing I noticed is you used your answer from Vb as your input voltage for your second divider equation. You should still use 6V as the supply for both equations. If you do that, you'll end up calculating answers that agree with your simulation.

I ended up with Vb = 5.1608V and Vc = 5.035V

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